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complex conjugate problem, already assumed real

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Dan Zhu
Dan Zhu on 6 Nov 2020
Commented: Raunak Gupta on 15 Dec 2020
I'm using the latest 2020b build of MATLAB. Here's part of my code:
syms nu theta H(r,theta)
assume([nu theta H(r,theta)], 'real')
simplify(sin(theta)'-sin(theta)) %This is 0.
simplify(diff(H(r,theta),theta)'-diff(H(r,theta),theta)) %This is not.
simplify(kroneckerDelta(nu)'-kroneckerDelta(nu)) %This is also not.
How to solve this?

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Answers (1)

Raunak Gupta
Raunak Gupta on 9 Nov 2020
Hi,
From the documentation of MATLAB Operators and Special Characters you can see that is used for complex conjugate transpose whereas . is used for transpose of a matrix. Since the first expression which gave zero is always real so both operators will behave same. In the other two expressions it may give complex number as output (based on the definition of nu and H(r , theta)). So, both operators are different in that case and thus you are not getting zero. If you replace the bottom two expression with the following it will give 0 as answer.
simplify(diff(H(r,theta),theta).'-diff(H(r,theta),theta))
simplify(kroneckerDelta(nu).'-kroneckerDelta(nu))

  2 Comments

Dan Zhu
Dan Zhu on 14 Dec 2020
That doesn't solve the problem, in my original code, there are other parts that are still complex in nature. I need to use complex conjugate, but I need those elements I've already assumed real to perform normally under the ' complex conjugate operator.
Raunak Gupta
Raunak Gupta on 15 Dec 2020
Hi,
For the real numbers, complex conjugate will return the same output as simple transpose. Since here you have both real and complex numbers in H(r,theta) the difference between both terms will not be an all zero matrix. For real numbers, matrix entries will be zero and for complex number it will be purely imaginary. Also make sure if diff(H(r,theta)) is a 2-D matrix it is symmetric or conjugate symmetric for the output to be zero.
Hope this clarifies!

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